For the galvanic (voltaic) cell Fe(s) + Mn²⁺(aq) ⟶ Fe²⁺(aq) + Mn(s) (E° = 0.77 V at 25 °C), what is [Fe²⁺] if [Mn²⁺] = 0.050 M and E = 0.78 V? Assume T is 298 K

Question

For the galvanic (voltaic) cell Fe(s) + Mn²⁺(aq) ⟶ Fe²⁺(aq) + Mn(s) (E° = 0.77 V at 25 °C), what is [Fe²⁺] if [Mn²⁺] = 0.050 M and E = 0.78 V? Assume T is 298 K

Answer 1

Fe(s) + Mn2+(aq)—> Fe2+(aq) + Mn(s) E°=0.77V

Fe lost two electrons and Mn gained two electrons, so number of electrons transferred is 2 (n=2).

Use Nernst equation:

Ecell = E°cell - (0.05916/n)logQ

0.78 = 0.77 - (0.05916/2)log([Fe2+]/[Mn2+])

0.01 = -0.02958log([Fe2+]/0.050)

-0.33807 = log([Fe2+]/0.050)

10^(-0.33807) = [Fe2+]/0.050

0.023 M = [Fe2+]